Question

A circular ring of radius $R$ and mass $m$ made of a uniform wire of cross-sectional area $A$ is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring. The breaking stress of the material of the ring is ${\sigma }_b$. Determine the maximum angular speed ${\omega }_{max}$ at which the ring may be rotated without failure.

• A. ${\omega }_{max}=\sqrt{\frac{\pi A{\sigma }_b}{mR}}$
• B. ${\omega }_{max}=\sqrt{\frac{\pi A{\sigma }_b}{2mR}}$
• C. ${\omega }_{max}=\sqrt{\frac{2\pi A{\sigma }_b}{mR}}$
• D. ${\omega }_{max}=\sqrt{\frac{4\pi A{\sigma }_b}{mR}}$

Answer 1

Answer: (C)

${\omega }_{max}=\sqrt{\frac{2\pi A{\sigma }_b}{mR}}$

Let the tension in the wire be $T$

Mass of the small element of the ring $dm=\frac{m}{2\pi }d\theta$

Net radial component of the tension $F^{\prime }=2F$ $sin\frac{d\theta }{2}$

As $d\theta$ is very small $⟹$ $sin\frac{d\theta }{2}\approx \frac{d\theta }{2}$

$⟹$ $F^{\prime }=Fd\theta$

Using $Fd\theta =\left(dm\right)R{\omega }^2$

OR $Fd\theta =\frac{m}{2\pi }d\theta \times R{\omega }^2$ $⟹F=\frac{mR{\omega }^2}{2\pi }$

Breaking stress ${\sigma }_b=\frac{F_{max}}{A}$

$\therefore$ ${\sigma }_b=\frac{mR{\omega }_{max}^2}{2\pi A}$ $⟹{\omega }_{max}=\sqrt{\frac{2\pi A{\sigma }_b}{mR}}$